ExamBro
ExamBro
JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)

\(50 \,mL\) of \(0.1\, M \,CH _{3} COOH\) is being titrated against \(0.1 \,M\, NaOH\). When \(25\, mL\) of \(NaOH\) has been added, the \(pH\) of the solution will be \(....\,\times 10^{-2}\). (Nearest integer) (Given : \(\left.pK _{ a }\left( CH _{3} COOH \right)=4.76\right)\) \(\log 2=0.30\)  \(\log 3=0.48\)  \(\log 5=0.69\)  \(\log 7=0.84\)  \(\log 11=1.04\)

  1. A \(963\)
  2. B \(123\)
  3. C \(476\)
  4. D \(596\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(476\)

Step-by-step Solution

Detailed explanation

Moles of \(CH _{3} COOH =5\, m\, mole\, moles\) of \(NaOH =2.5\, m\, mole\) \(NaOH + CH _{3} COOH \longrightarrow CH _{3} COONa + H _{2} O\) \(2.5\, m\, mole \quad\quad 2.5 \,m\,mole\) \(0 \quad\quad\quad\quad\quad 2.5 m \text { mole } \quad\quad2.5 m \text { mole }\) so buffer…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app